How are patterns used to create algebraic expressions?

Question

Eva Abramson · Accepted Answer

undefined Patterns are used to create algebraic expressions by identifying the relationship between the variables involved in the pattern and expressing this relationship using mathematical symbols and operations. This often involves observing how the values of certain variables change or relate to each other within the pattern, and then using this information to construct an algebraic expression that represents the pattern. This expression may include variables, constants, and arithmetic operations such as addition, subtraction, multiplication, and division. By analyzing patterns and translating them into algebraic expressions, mathematicians and scientists can model real-world phenomena, make predictions, and solve problems efficiently.

Abigail Allen · Answer

This question is rather general, so I will only address a small fragment of it...

How can patterns be recognized and expressed using algebraic expressions? For example, given the sequence #1, 4, 10, 20, 35, 56# What is the pattern? How do you find the next number in the sequence? What is the formula for the #n^(th)# term in the sequence? When dealing with these kinds of issues, it's usually beneficial to create a series of distinctions between succeeding terms. Keep doing this until the sequence becomes constant. #1, 4, 10, 20, 35, 56# # -> 3, 6, 10, 15, 21# # -> 3, 4, 5, 6 # # -> 1, 1, 1 # Since it has taken #3# steps to get to a constant sequence, the original sequence is expressible as a cubic expression. We can directly construct the formula for #a_n# from the first term of each of these sequences as: #a_n = color(red)(1) + color(red)(3) * n/(1!) + color(red)(3) * (n(n-1))/(2!) + color(red)(1) * (n(n-1)(n-2))/(3!)# This discrete version of Taylor's theorem appeals to me. With the Fibonacci sequence, which is essentially exponential rather than polynomial, this method will not work well. #0, 1, 1, 2, 3, 5, 8, 13, 21, 34# #-> 1, 0, 1, 1, 2, 3, 5, 8, 13# #-> -1, 1, 0, 1, 1, 2, 3, 5# #-> 2, -1, 1, 0, 1, 1, 2# #-> -3, 2, -1, 1, 0, 1# #-> 5, -3, 2, -1, 1# ... But it is possible to find a formula for #F_n#, viz #F_n = (phi^n - (-phi)^-n)/sqrt(5)# where #phi = (1+sqrt(5))/2# Not even the field of numerical sequences can be fully addressed.